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[DOTOANNANG]

$a, b> 0$, $ab= 1$. CM:

$$\frac{1}{a^{3}+ b^{2}+ 1}+ \frac{1}{a^{2}+ 2}+ \frac{1}{b^{3}+ 2}\leq 1$$

[PugMath]

$a, b> 0$, $ab= 1$. CM:

$$\frac{1}{a^{3}+ b^{2}+ 1}+ \frac{1}{a^{2}+ 2}+ \frac{1}{b^{3}+ 2}\leq 1$$

$taco:(a^{3}+b^{2}+1)(\frac{1}{a}+1+1)\geqslant (a+b+1)^{2} ;(a^{2}+1+1)(1+b^{2}+1)\geqslant (a+b+1)^2;(b^{3}+1+1)(\frac{1}{b}+a^{2}+1)\geqslant (a+b+1)^{2} =>\frac{1}{a^{3}+b^{2}+1}+\frac{1}{a^{2}+2}+\frac{1}{b^3+2}\leqslant \frac{\frac{1}{a}+2+2+b^{2}+\frac{1}{b}+a^{2}+1}{(a+b+1)^{2}}=\frac{a^{2}+b^{2}+a+b+5}{(a+b+1)^{2}}=\frac{a^{2}+b^{2}+2a+2b+2ab+1-a-b-1+3}{(a+b+1)^{2}}=1-\frac{1}{(a+b+1)}+\frac{3}{(a+b+1)^{2}} tudo.can.c/m-\frac{1}{(a+b+1)}+\frac{3}{(a+b+1)^{2}}\leqslant 0 <=>-a-b-1+3\leqslant 0<=>-a-b\leqslant -2=>a+b \geqslant 2.taco:(a+b)^{2}\geqslant 4ab=4=>a+b\geqslant 2(Q.E.D)$